Integrand size = 31, antiderivative size = 261 \[ \int \sec ^m(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {b (A b (2+m)+a B (3+m)) \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+m) (2+m)}+\frac {b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{d (2+m)}-\frac {\left (A b^2 m+2 a b B m+a^2 A (1+m)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d \left (1-m^2\right ) \sqrt {\sin ^2(c+d x)}}+\frac {\left (b^2 B (1+m)+a (2 A b+a B) (2+m)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},\frac {2-m}{2},\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d m (2+m) \sqrt {\sin ^2(c+d x)}} \]
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Time = 0.29 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4111, 4132, 3857, 2722, 4131} \[ \int \sec ^m(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=-\frac {\sin (c+d x) \left (a^2 A (m+1)+2 a b B m+A b^2 m\right ) \sec ^{m-1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(c+d x)\right )}{d \left (1-m^2\right ) \sqrt {\sin ^2(c+d x)}}+\frac {\sin (c+d x) \left (a (m+2) (a B+2 A b)+b^2 B (m+1)\right ) \sec ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},\frac {2-m}{2},\cos ^2(c+d x)\right )}{d m (m+2) \sqrt {\sin ^2(c+d x)}}+\frac {b \sin (c+d x) (a B (m+3)+A b (m+2)) \sec ^{m+1}(c+d x)}{d (m+1) (m+2)}+\frac {b B \sin (c+d x) \sec ^{m+1}(c+d x) (a+b \sec (c+d x))}{d (m+2)} \]
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Rule 2722
Rule 3857
Rule 4111
Rule 4131
Rule 4132
Rubi steps \begin{align*} \text {integral}& = \frac {b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{d (2+m)}+\frac {\int \sec ^m(c+d x) \left (a (b B m+a A (2+m))+\left (b^2 B (1+m)+a (2 A b+a B) (2+m)\right ) \sec (c+d x)+b (A b (2+m)+a B (3+m)) \sec ^2(c+d x)\right ) \, dx}{2+m} \\ & = \frac {b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{d (2+m)}+\frac {\int \sec ^m(c+d x) \left (a (b B m+a A (2+m))+b (A b (2+m)+a B (3+m)) \sec ^2(c+d x)\right ) \, dx}{2+m}+\left (2 a A b+a^2 B+\frac {b^2 B (1+m)}{2+m}\right ) \int \sec ^{1+m}(c+d x) \, dx \\ & = \frac {b (A b (2+m)+a B (3+m)) \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+m) (2+m)}+\frac {b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{d (2+m)}+\frac {\left (A b^2 m+2 a b B m+a^2 A (1+m)\right ) \int \sec ^m(c+d x) \, dx}{1+m}+\left (\left (2 a A b+a^2 B+\frac {b^2 B (1+m)}{2+m}\right ) \cos ^m(c+d x) \sec ^m(c+d x)\right ) \int \cos ^{-1-m}(c+d x) \, dx \\ & = \frac {b (A b (2+m)+a B (3+m)) \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+m) (2+m)}+\frac {b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{d (2+m)}+\frac {\left (2 a A b+a^2 B+\frac {b^2 B (1+m)}{2+m}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},\frac {2-m}{2},\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d m \sqrt {\sin ^2(c+d x)}}+\frac {\left (\left (A b^2 m+2 a b B m+a^2 A (1+m)\right ) \cos ^m(c+d x) \sec ^m(c+d x)\right ) \int \cos ^{-m}(c+d x) \, dx}{1+m} \\ & = \frac {b (A b (2+m)+a B (3+m)) \sec ^{1+m}(c+d x) \sin (c+d x)}{d (1+m) (2+m)}+\frac {b B \sec ^{1+m}(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{d (2+m)}-\frac {\left (A b^2 m+2 a b B m+a^2 A (1+m)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-m}{2},\frac {3-m}{2},\cos ^2(c+d x)\right ) \sec ^{-1+m}(c+d x) \sin (c+d x)}{d \left (1-m^2\right ) \sqrt {\sin ^2(c+d x)}}+\frac {\left (2 a A b+a^2 B+\frac {b^2 B (1+m)}{2+m}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {m}{2},\frac {2-m}{2},\cos ^2(c+d x)\right ) \sec ^m(c+d x) \sin (c+d x)}{d m \sqrt {\sin ^2(c+d x)}} \\ \end{align*}
Time = 1.09 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.92 \[ \int \sec ^m(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {\csc (c+d x) \left (a^2 A \left (6+11 m+6 m^2+m^3\right ) \cos ^3(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(c+d x)\right )+a (2 A b+a B) m \left (6+5 m+m^2\right ) \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\sec ^2(c+d x)\right )+b m (1+m) \left ((A b+2 a B) (3+m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\sec ^2(c+d x)\right )+b B (2+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+m}{2},\frac {5+m}{2},\sec ^2(c+d x)\right )\right )\right ) \sec ^{2+m}(c+d x) \sqrt {-\tan ^2(c+d x)}}{d m (1+m) (2+m) (3+m)} \]
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\[\int \sec \left (d x +c \right )^{m} \left (a +b \sec \left (d x +c \right )\right )^{2} \left (A +B \sec \left (d x +c \right )\right )d x\]
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\[ \int \sec ^m(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{m} \,d x } \]
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\[ \int \sec ^m(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int \left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sec ^{m}{\left (c + d x \right )}\, dx \]
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\[ \int \sec ^m(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{m} \,d x } \]
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\[ \int \sec ^m(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{m} \,d x } \]
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Timed out. \[ \int \sec ^m(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\int \left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m \,d x \]
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